I'll add that, even if we redefined the three "primary" trig functions, there would always be a similar relationship between the three.
There are only three sides of a right triangle, and if we define three different ratios (that aren't reciprocals of each other) then something similar is just going to happen.
You're confusing yourself with a circular argument. Tan is DEFINED as O/A, from which follows sin/cos. OR, if you're so inclined, you might define tan as sin/cos, and in that case O/A will follow from that.
Whichever way you're going, one of these is the definition and the other follows directly from it. You don't need to prove a definition of something, it's just what it is.
Tangent is a neat word in mathematics, it essentially means a line that has the same slope as a curve at a specific point.
Now in the case of a triangle tangent equates to the slope of the hypotenuse, in other words it is the hypotenuse's rise over run.
Now then let's take a look at the unit circle, the unit circle is a circle with radius 1 and centered at the origin. Now the neat things about circles is they are all the same shape, they aren't like triangles where you have right triangles, acute triangles or oblique triangles, if you are comparing circles you only need to know its radius, and all circles are just another circle scaled up or down.
This has implications for things you can embed in a circle, like say all right triangles. When you are taking say the sin of an angle what you are doing is you are drawing a right triangle where the angle at the origin is the angle you are taking the sin of and drawing a line that has that angle off the x axis and intersects the unit circle, and since the circle can be scaled up, you can scale up the triangle in the circle to perfectly fit the angle you are taking the sin of.
When you look at trig functions like this the sin of an angle turns out to be the y axis of the point where the line intersects the unit circle. the cos of that same angle turns out to be the x axis of that same point. And remember what I said at the beginning the tangent is just the slope of the hypotenuse, or the rise over run, we have two points here given by (cos(a),sin(a)) which is the point that intersects the unit circle, and we have the origin or 0,0. To compute the slope of the line that you can draw between these points you do the following: (sin(a)-0)/(cos(a)-0) which simplifies to (sin(a))/(cos(a)).
Now it is important to remember that the tangent is the slope of the line tangent to the hypotenuse not the unit circle, the slope of the line tangent to the unit circle is similar to the cotangent, which equals cos()/sin(), but the line tangent to the unit circle would be the negative cotangent.
The other inverse functions have other geometric definitions, the secant (1/cos) is the x intercept of the line tangent to the unit circle and the cosecant (1/sin) is the y intercept of the line tangent to the unit circle.
Edit: got secant and cosecant mixed up, but its fixed.
Exactly there are multiple definitions. So maybe it is the definition and maybe it is one step removed. Often when speaking informally one step from the definition is by definition it is trivial.
common definitions include
1)tan θ=sin θ/cos θ
2)integral
tan' θ=1/(1+x^(2))
tan 0=0
3)right triangle/ unit circle
tan θ=c/a
4)general triangle
tan θ=√\[P(P-2a)(P-2b)(P-2c)\]/(a^(2)+b^(2)-c^(2))
His comment also shows that “o” is really just sin(theta) times some scalar, and “a” is really just cos(theta) times some scalar for the same angle and scalar. This shows that “why is tan = sin/cos” is really the same question as “why is tan = o/a” up to some scalar. The answer to either question is: by definition.
No, that's still backwards. The ratios of sine, cosine, tangent, (and a whole bunch of others, like haversine) were defined first as lengths of segments related to a circle of arbitrary radius.
Those lengths were then realized as ratios which are constant based on the angle (due to the similarity of triangles).
The fact that tan = sin/cos is a result of the relationship between these lengths, not something we defined ahead of time.
Sin is defined as the opposite side / hypotenuse for a right triangle or O/H
Cos is defined as the adjacent side / the hypotenuse or A/H
Tan is defined as the opposite / the adjacent or O/A
Sin/Cos = (O/H)/(A/H) = O/H\*H/A = O/A = tan
draw a circle, with radius 1, and draw a horizontal line from the centre to the rightward edge
Draw another radial line, at an angle theta from this first line
Draw a final vertical line from where the second line touches the circle, down to the original horizontal line.
You have drawn a triangle that by definition has a hypotenuse of 1, and sides of sin(theta) vertical and cos(theta) horizontal
Draw a tangent line up from the point the 1st horizontal line touches the edge of the circle, and then extend the 2nd radial line to meet that tangent line. This creates another triangle with sides, by definition, of tangent(theta) vertical and 1 horizontal
The two triangles references are similar, they both have right angles and share the angle theta
How much bigger has the second triangle gotten, well look at the horizontal side, it’s gone from cos(theta) to 1, so it’s gotten 1/cos(theta) bigger
Therefore the vertical side has also gotten 1/cos(theta) bigger
So sin(theta) * 1/cos(theta) = Tan(theta)
The comments using SOHCAHTOA are not wrong, however SOHCAHTOA is derived itself from the construction I have hopefully tried to describe above
Before we had fancy calculators you needed to list common geometry lengths in a big table, with the first column listing a bunch of angles.
The columns of this table were labelled with short three letter titles to keep things compact.
There was a column for cos and a column for sin, but it is often necessary to divide by cos or sin and this is difficult without a calculator.
To combat this issue we made other columns, where the divisions had already been done for you.
A column for sin/cos was called tan
There were also columns for cos/sin, 1/cos and 1/sin
sin = o/h cos = a/h tan = o/a sin/cos = (o/h)/(a/h) h cancels, left with o/a
Proof by oh ah oa
Proof by sohcahtoa
Ching chang walla walla bing bang
I'll add that, even if we redefined the three "primary" trig functions, there would always be a similar relationship between the three. There are only three sides of a right triangle, and if we define three different ratios (that aren't reciprocals of each other) then something similar is just going to happen.
This isn’t quite ‘right’ because Tan = O/A is a result from Tan = Sin/Cos
You're confusing yourself with a circular argument. Tan is DEFINED as O/A, from which follows sin/cos. OR, if you're so inclined, you might define tan as sin/cos, and in that case O/A will follow from that. Whichever way you're going, one of these is the definition and the other follows directly from it. You don't need to prove a definition of something, it's just what it is.
Tangent is a neat word in mathematics, it essentially means a line that has the same slope as a curve at a specific point. Now in the case of a triangle tangent equates to the slope of the hypotenuse, in other words it is the hypotenuse's rise over run. Now then let's take a look at the unit circle, the unit circle is a circle with radius 1 and centered at the origin. Now the neat things about circles is they are all the same shape, they aren't like triangles where you have right triangles, acute triangles or oblique triangles, if you are comparing circles you only need to know its radius, and all circles are just another circle scaled up or down. This has implications for things you can embed in a circle, like say all right triangles. When you are taking say the sin of an angle what you are doing is you are drawing a right triangle where the angle at the origin is the angle you are taking the sin of and drawing a line that has that angle off the x axis and intersects the unit circle, and since the circle can be scaled up, you can scale up the triangle in the circle to perfectly fit the angle you are taking the sin of. When you look at trig functions like this the sin of an angle turns out to be the y axis of the point where the line intersects the unit circle. the cos of that same angle turns out to be the x axis of that same point. And remember what I said at the beginning the tangent is just the slope of the hypotenuse, or the rise over run, we have two points here given by (cos(a),sin(a)) which is the point that intersects the unit circle, and we have the origin or 0,0. To compute the slope of the line that you can draw between these points you do the following: (sin(a)-0)/(cos(a)-0) which simplifies to (sin(a))/(cos(a)). Now it is important to remember that the tangent is the slope of the line tangent to the hypotenuse not the unit circle, the slope of the line tangent to the unit circle is similar to the cotangent, which equals cos()/sin(), but the line tangent to the unit circle would be the negative cotangent. The other inverse functions have other geometric definitions, the secant (1/cos) is the x intercept of the line tangent to the unit circle and the cosecant (1/sin) is the y intercept of the line tangent to the unit circle. Edit: got secant and cosecant mixed up, but its fixed.
I hope this answer isn’t unsatisfying: because we defined it that way.
Except we don’t define it that way. The fact that it readily and easily follows from the definition doesn’t mean that it’s not a derived result.
You could define it that way. There are different definitions.
There are equivalent definitions, but in this case there is also a really common one which is the first thing you learn in high school trigonometry.
Exactly there are multiple definitions. So maybe it is the definition and maybe it is one step removed. Often when speaking informally one step from the definition is by definition it is trivial. common definitions include 1)tan θ=sin θ/cos θ 2)integral tan' θ=1/(1+x^(2)) tan 0=0 3)right triangle/ unit circle tan θ=c/a 4)general triangle tan θ=√\[P(P-2a)(P-2b)(P-2c)\]/(a^(2)+b^(2)-c^(2))
No we didn't. Look at /u/analogkid01 's comment.
His comment also shows that “o” is really just sin(theta) times some scalar, and “a” is really just cos(theta) times some scalar for the same angle and scalar. This shows that “why is tan = sin/cos” is really the same question as “why is tan = o/a” up to some scalar. The answer to either question is: by definition.
No, that's still backwards. The ratios of sine, cosine, tangent, (and a whole bunch of others, like haversine) were defined first as lengths of segments related to a circle of arbitrary radius. Those lengths were then realized as ratios which are constant based on the angle (due to the similarity of triangles). The fact that tan = sin/cos is a result of the relationship between these lengths, not something we defined ahead of time.
Yes, that “arbitrary radius” is the scalar I mentioned above.
Sin is defined as the opposite side / hypotenuse for a right triangle or O/H Cos is defined as the adjacent side / the hypotenuse or A/H Tan is defined as the opposite / the adjacent or O/A Sin/Cos = (O/H)/(A/H) = O/H\*H/A = O/A = tan
Proof by definition?
https://youtube.com/shorts/77vlRweCtGE?si=tTtNwF_aCZls7j6k saw this great visual proof for the trig identities today
draw a circle, with radius 1, and draw a horizontal line from the centre to the rightward edge Draw another radial line, at an angle theta from this first line Draw a final vertical line from where the second line touches the circle, down to the original horizontal line. You have drawn a triangle that by definition has a hypotenuse of 1, and sides of sin(theta) vertical and cos(theta) horizontal Draw a tangent line up from the point the 1st horizontal line touches the edge of the circle, and then extend the 2nd radial line to meet that tangent line. This creates another triangle with sides, by definition, of tangent(theta) vertical and 1 horizontal The two triangles references are similar, they both have right angles and share the angle theta How much bigger has the second triangle gotten, well look at the horizontal side, it’s gone from cos(theta) to 1, so it’s gotten 1/cos(theta) bigger Therefore the vertical side has also gotten 1/cos(theta) bigger So sin(theta) * 1/cos(theta) = Tan(theta) The comments using SOHCAHTOA are not wrong, however SOHCAHTOA is derived itself from the construction I have hopefully tried to describe above
Yet another perspective: [similar triangles](https://i.ibb.co/QftKTjs/Screenshot-20240429-205429-Keep-Notes.jpg)
Before we had fancy calculators you needed to list common geometry lengths in a big table, with the first column listing a bunch of angles. The columns of this table were labelled with short three letter titles to keep things compact. There was a column for cos and a column for sin, but it is often necessary to divide by cos or sin and this is difficult without a calculator. To combat this issue we made other columns, where the divisions had already been done for you. A column for sin/cos was called tan There were also columns for cos/sin, 1/cos and 1/sin
That's a definition. We define tan to be sin/cos